Exercise 2.1: Sets and Functions
1. Express the following sets in the set builder notation:
(1) $\{x \in \mathbb{N} : x^2 > 9 \text{ and } x^3 < 130\}$
Solution: Here, the set of natural numbers is $\mathbb{N} = \{1, 2, 3, 4, 5, 6, \dots\}$
Now,
For $x = 1$, $x^2 = 1^2 = 1 \ngtr 9$ and $x^3 = 1^3 = 1 < 130$
For $x = 2$, $x^2 = 2^2 = 4 \ngtr 9$ and $x^3 = 2^3 = 8 < 130$
For $x = 3$, $x^2 = 3^2 = 9 \ngtr 9$ and $x^3 = 3^3 = 27 < 130$
For $x = 4$, $x^2 = 4^2 = 16 > 9$ and $x^3 = 4^3 = 64 < 130$
For $x = 5$, $x^2 = 5^2 = 25 > 9$ and $x^3 = 5^3 = 125 < 130$
For $x = 6$, $x^2 = 6^2 = 36 > 9$ and $x^3 = 6^3 = 216 \nless 130$
According to the given conditions, the acceptable numbers are: $4, 5$
Answer:$\{4, 5\}$
(2) $\{x \in \mathbb{Z} : x^2 > 5 \text{ and } x^3 \le 36\}$
Solution: Here, the set of integers is $\mathbb{Z} = \{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}$
Checking positive and negative values:
For $x = 0$, $x^2 = 0 \ngtr 5$ and $x^3 = 0 \le 36$
For $x = 1$, $x^2 = 1 \ngtr 5$ and $x^3 = 1 \le 36$
For $x = 2$, $x^2 = 4 \ngtr 5$ and $x^3 = 8 \le 36$
For $x = 3$, $x^2 = 9 > 5$ and $x^3 = 27 \le 36$
For $x = 4$, $x^2 = 16 > 5$ and $x^3 = 64 \nleq 36$
For negative numbers:
For $x = -1$, $x^2 = 1 \ngtr 5$ and $x^3 = -1 \le 36$
For $x = -2$, $x^2 = 4 \ngtr 5$ and $x^3 = -8 \le 36$
For $x = -3$, $x^2 = 9 > 5$ and $x^3 = -27 \le 36$
For $x = -4$, $x^2 = 16 > 5$ and $x^3 = -64 \le 36$
Similarly, for all negative values of $x \le -3$, both conditions $x^2 > 5$ and $x^3 \le 36$ will hold true.
According to the conditions, the acceptable numbers are: $3, -3, -4, -5, \dots$
Answer:$\{\dots, -5, -4, -3, 3\}$
(3) $\{x \in \mathbb{N} : x \text{ is a factor of } 36 \text{ and a multiple of } 6\}$
Solution: Factors of $36$: $1, 2, 3, 4, 6, 9, 12, 18, 36$
Multiples of $6$: $6, 12, 18, 24, 30, 36, 42, \dots$
The numbers that satisfy both conditions are: $6, 12, 18, 36$
Answer:$\{6, 12, 18, 36\}$
(4) $\{x \in \mathbb{N} : x^3 > 25 \text{ and } x^4 < 264\}$
Solution: Set of natural numbers, $\mathbb{N} = \{1, 2, 3, 4, 5, \dots\}$
For $x = 1$, $x^3 = 1 \ngtr 25$
For $x = 2$, $x^3 = 8 \ngtr 25$
For $x = 3$, $x^3 = 27 > 25$ and $x^4 = 81 < 264$
For $x = 4$, $x^3 = 64 > 25$ and $x^4 = 256 < 264$
For $x = 5$, $x^3 = 125 > 25$ and $x^4 = 625 \nless 264$
According to the conditions, the acceptable numbers are: $3, 4$
Answer:$\{3, 4\}$
2. Express the following sets in the set builder notation:
(1) $\{3, 5, 7, 9, 11\}$
Solution: The elements are odd numbers greater than $1$ and less than $13$.
Answer:$\{x \in \mathbb{N} : x \text{ is an odd number and } 1 < x < 13\}$
(2) $\{1, 2, 3, 4, 6, 9, 12, 18, 36\}$
Solution: Each of the elements is a factor of $36$.
Answer:$\{x \in \mathbb{N} : x \text{ is a factor of } 36\}$
(3) $\{4, 8, 12, 16, 20, 24, 28, 32, 36, 40\}$
Solution: The elements are multiples of $4$ and less than or equal to $40$.
Answer:$\{x \in \mathbb{N} : x \text{ is a multiple of } 4 \text{ and } x \le 40\}$
(4) $\{\pm 4, \pm 5, \pm 6\}$
Solution: Here, the elements are integers ($\mathbb{Z}$).
Squaring them gives: $4^2 = 16$, $5^2 = 25$, $6^2 = 36$
Cubing them gives: $4^3 = 64$, $5^3 = 125$, $6^3 = 216$
According to the conditions, the square is greater than $9$ and the cube is less than or equal to $216$.
Answer:$\{x \in \mathbb{Z} : x^2 > 9 \text{ and } x^3 \le 216\}$
3. If $A = \{2, 3, 4\}$, $B = \{1, 2, a\}$, and $C = \{2, a, b\}$, then determine the following sets:
(1) $B \setminus C$
Solution: $B \setminus C = \{1, 2, a\} \setminus \{2, a, b\} = \{1\}$
Answer:$\{1\}$
(2) $A \cup B$
Solution: $A \cup B = \{2, 3, 4\} \cup \{1, 2, a\} = \{1, 2, 3, 4, a\}$
Answer:$\{1, 2, 3, 4, a\}$
(3) $A \cap C$
Solution: $A \cap C = \{2, 3, 4\} \cap \{2, a, b\} = \{2\}$
Answer:$\{2\}$
(4) $A \cup (B \cap C)$
Solution: First, $B \cap C = \{1, 2, a\} \cap \{2, a, b\} = \{2, a\}$
Now, $A \cup (B \cap C) = \{2, 3, 4\} \cup \{2, a\} = \{2, 3, 4, a\}$
Answer:$\{2, 3, 4, a\}$
(5) $A \cap (B \cup C)$
Solution: First, $B \cup C = \{1, 2, a\} \cup \{2, a, b\} = \{1, 2, a, b\}$
Now, $A \cap (B \cup C) = \{2, 3, 4\} \cap \{1, 2, a, b\} = \{2\}$
Answer:$\{2\}$
4. If $U = \{1, 2, 3, 4, 5, 6, 7\}$, $A = \{1, 3, 5\}$, $B = \{2, 4, 6\}$, and $C = \{3, 4, 5, 6, 7\}$, verify the following statements:
(1) $(A \cup B)’ = A’ \cap B’$
Solution: L.H.S. $= A \cup B = \{1, 3, 5\} \cup \{2, 4, 6\} = \{1, 2, 3, 4, 5, 6\}$
$\therefore (A \cup B)’ = U \setminus (A \cup B) = \{1, 2, 3, 4, 5, 6, 7\} \setminus \{1, 2, 3, 4, 5, 6\} = \{7\}$
R.H.S.:
$A’ = U \setminus A = \{1, 2, 3, 4, 5, 6, 7\} \setminus \{1, 3, 5\} = \{2, 4, 6, 7\}$
$B’ = U \setminus B = \{1, 2, 3, 4, 5, 6, 7\} \setminus \{2, 4, 6\} = \{1, 3, 5, 7\}$
$\therefore A’ \cap B’ = \{2, 4, 6, 7\} \cap \{1, 3, 5, 7\} = \{7\}$
L.H.S. = R.H.S.
(2) $(B \cap C)’ = B’ \cup C’$
Solution: L.H.S.:
$B \cap C = \{2, 4, 6\} \cap \{3, 4, 5, 6, 7\} = \{4, 6\}$
$\therefore (B \cap C)’ = U \setminus (B \cap C) = \{1, 2, 3, 4, 5, 6, 7\} \setminus \{4, 6\} = \{1, 2, 3, 5, 7\}$
R.H.S.:
$B’ = \{1, 3, 5, 7\}$
$C’ = U \setminus C = \{1, 2, 3, 4, 5, 6, 7\} \setminus \{3, 4, 5, 6, 7\} = \{1, 2\}$
$\therefore B’ \cup C’ = \{1, 3, 5, 7\} \cup \{1, 2\} = \{1, 2, 3, 5, 7\}$
L.H.S. = R.H.S.
(3) $(A \cup B) \cap C = (A \cap C) \cup (B \cap C)$
Solution: L.H.S.:
$A \cup B = \{1, 2, 3, 4, 5, 6\}$
$\therefore (A \cup B) \cap C = \{1, 2, 3, 4, 5, 6\} \cap \{3, 4, 5, 6, 7\} = \{3, 4, 5, 6\}$
Right-Hand Side (RHS):
$A \cap C = \{1, 3, 5\} \cap \{3, 4, 5, 6, 7\} = \{3, 5\}$
$B \cap C = \{2, 4, 6\} \cap \{3, 4, 5, 6, 7\} = \{4, 6\}$
$\therefore (A \cap C) \cup (B \cap C) = \{3, 5\} \cup \{4, 6\} = \{3, 4, 5, 6\}$
L.H.S. = R.H.S.
(4) $(A \cap B) \cup C = (A \cup C) \cap (B \cup C)$
Solution:
L.H.S.:
$A \cap B = \{1, 3, 5\} \cap \{2, 4, 6\} = \emptyset$
$\therefore (A \cap B) \cup C = \emptyset \cup \{3, 4, 5, 6, 7\} = \{3, 4, 5, 6, 7\}$
R.H.S.:
$A \cup C = \{1, 3, 5\} \cup \{3, 4, 5, 6, 7\} = \{1, 3, 4, 5, 6, 7\}$
$B \cup C = \{2, 4, 6\} \cup \{3, 4, 5, 6, 7\} = \{2, 3, 4, 5, 6, 7\}$
$\therefore (A \cup C) \cap (B \cup C) = \{1, 3, 4, 5, 6, 7\} \cap \{2, 3, 4, 5, 6, 7\} = \{3, 4, 5, 6, 7\}$
L.H.S. = R.H.S.
5. If $Q = \{x, y\}$ and $R = \{m, n, l\}$, then determine $P(Q)$ and $P(R)$.
Solution: Given, $Q = \{x, y\}$
$\therefore P(Q) = \{\{x, y\}, \{x\}, \{y\}, \emptyset\}$
Also,
$R = \{m, n, l\}$
$\therefore P(R) = \{\{m, n, l\}, \{m, n\}, \{m, l\}, \{n, l\}, \{m\}, \{n\}, \{l\}, \emptyset\}$
Answer:
$P(Q) = \{\{x, y\}, \{x\}, \{y\}, \emptyset\}$
$P(R) = \{\{m, n, l\}, \{m, n\}, \{m, l\}, \{n, l\}, \{m\}, \{n\}, \{l\}, \emptyset\}$
6. If $A = \{a, b\}$, $B = \{a, b, c\}$, and $C = A \cup B$, then show that the number of elements of $P(C)$ is $2^n$, where $n$ is the number of elements of $C$.
Solution: Given,
$A = \{a, b\}$
$B = \{a, b, c\}$
$\therefore C = A \cup B = \{a, b\} \cup \{a, b, c\} = \{a, b, c\}$
Here, the number of elements of $C$ is $n = 3$.
Now, the power set of $C$:
$P(C) = \{\{a, b, c\}, \{a, b\}, \{a, c\}, \{b, c\}, \{a\}, \{b\}, \{c\}, \emptyset\}$
$\therefore$ Number of elements of $P(C) = 8 = 2^3 = 2^n$ (Shown)
7. (1) If $(x – 1, y + 2) = (y – 2, 2x + 1)$, then determine $x$ and $y$.
Solution: By the definition of ordered pairs,
$x – 1 = y – 2 \implies x – y = -2 + 1 \implies x – y = -1 \text{ — (i)}$
and
$y + 2 = 2x + 1 \implies 2x – y = 2 – 1 \implies 2x – y = 1 \text{ — (ii)}$
Subtracting equation (i) from equation (ii):
$(2x – y) – (x – y) = 1 – (-1)$
$\implies x = 2$
Substituting the value $x = 2$ into equation (i):
$2 – y = -1 \implies y = 3$
Answer:$x = 2, y = 3$
(2) If $(ax – cy, a^2 – c^2) = (0, ay – cx)$, then determine $(x, y)$.
Solution: By the definition of ordered pairs,
$ax – cy = 0 \implies ax = cy \implies x = \frac{cy}{a} \text{ — (i)}$
and
$ay – cx = a^2 – c^2 \text{ — (ii)}$
Substituting the value of $x$ from equation (i) into equation (ii):
$ay – c\left(\frac{cy}{a}\right) = a^2 – c^2$
$\implies \frac{a^2 y – c^2 y}{a} = a^2 – c^2$
$\implies y(a^2 – c^2) = a(a^2 – c^2)$
$\implies y = a$
Substituting $y = a$ into equation (i):
$x = \frac{c \cdot a}{a} = c$
Answer:$(x, y) = (c, a)$
(3) If $(6x – y, 13) = (1, 3x + 2y)$, then determine $(x, y)$.
Solution: By the definition of ordered pairs,
$6x – y = 1 \implies y = 6x – 1 \text{ — (i)}$
and
$3x + 2y = 13 \text{ — (ii)}$
Substituting the value of $y$ from equation (i) into equation (ii):
$3x + 2(6x – 1) = 13$
$\implies 3x + 12x – 2 = 13$
$\implies 15x = 15$
$\implies x = 1$
Substituting $x = 1$ into equation (i):
$y = 6(1) – 1 = 5$
Answer:$(x, y) = (1, 5)$
8. (1) If $P = \{a\}$ and $Q = \{b, c\}$, then determine $P \times Q$ and $Q \times P$.
Solution: Given, $P = \{a\}$ and $Q = \{b, c\}$
$P \times Q = \{a\} \times \{b, c\} = \{(a, b), (a, c)\}$
and
$Q \times P = \{b, c\} \times \{a\} = \{(b, a), (c, a)\}$
Answer:$P \times Q = \{(a, b), (a, c)\}$ and $Q \times P = \{(b, a), (c, a)\}$
(2) If $A = \{3, 4, 5\}$, $B = \{4, 5, 6\}$, and $C = \{x, y\}$, determine $(A \cap B) \times C$.
Solution: Given, $A = \{3, 4, 5\}$, $B = \{4, 5, 6\}$, and $C = \{x, y\}$
First,
$A \cap B = \{3, 4, 5\} \cap \{4, 5, 6\} = \{4, 5\}$
$\therefore (A \cap B) \times C = \{4, 5\} \times \{x, y\} = \{(4, x), (4, y), (5, x), (5, y)\}$
Answer:$\{(4, x), (4, y), (5, x), (5, y)\}$
(3) If $P = \{3, 5, 7\}$, $Q = \{5, 7\}$, and $R = P \setminus Q$, determine $(P \cup Q) \times R$.
Solution: Given, $P = \{3, 5, 7\}$ and $Q = \{5, 7\}$
Here, $R = P \setminus Q = \{3, 5, 7\} \setminus \{5, 7\} = \{3\}$
and
$P \cup Q = \{3, 5, 7\} \cup \{5, 7\} = \{3, 5, 7\}$
$\therefore (P \cup Q) \times R = \{3, 5, 7\} \times \{3\} = \{(3, 3), (5, 3), (7, 3)\}$
Answer: {(3, 3), (5, 3), (7, 3)}
9. If $A$ and $B$ are respectively sets of all factors of $35$ and $45$ respectively, then determine $A \cup B$ and $A \cap B$.
Solution: All factors of $35$: $1, 5, 7, 35$
$\therefore A = \{1, 5, 7, 35\}$
All factors of $45$: $1, 3, 5, 9, 15, 45$
$\therefore B = \{1, 3, 5, 9, 15, 45\}$
Now,
$A \cup B = \{1, 5, 7, 35\} \cup \{1, 3, 5, 9, 15, 45\} = \{1, 3, 5, 7, 9, 15, 35, 45\}$
and
$A \cap B = \{1, 5, 7, 35\} \cap \{1, 3, 5, 9, 15, 45\} = \{1, 5\}$
Answer:$A \cup B = \{1, 3, 5, 7, 9, 15, 35, 45\}$ and $A \cap B = \{1, 5\}$
10. Determine the natural numbers that divided both $346$ and $556$ to have $31$ as residue .
Solution: The number that leaves a remainder of $31$ when dividing $346$ and $556$ must be greater than $31$ and must be a common factor of $(346 – 31)$ and $(556 – 31)$.
Here,
$346 – 31 = 315$
$556 – 31 = 525$
Let $A$ be the set of factors of $315$ greater than $31$, and $B$ be the set of factors of $525$ greater than $31$.
All factors of $315$: $1, 3, 5, 7, 9, 15, 21, 35, 45, 63, 105, 315$
$\therefore$ Set of factors greater than $31$, $A = \{35, 45, 63, 105, 315\}$
All factors of $525$: $1, 3, 5, 7, 15, 21, 25, 35, 75, 105, 175, 525$
$\therefore$ Set of factors greater than $31$, $B = \{35, 75, 105, 175, 525\}$
The required set will be $A \cap B$:
$A \cap B = \{35, 45, 63, 105, 315\} \cap \{35, 75, 105, 175, 525\} = \{35, 105\}$
Answer:$\{35, 105\}$
11. Out of $30$ students in a class, $20$ like football and $15$ like cricket. The number of students who like both games is $10$. Using Venn diagram determine the number of students who do not like either of the games.
Solution: Let, Set of total students = $U$, $\therefore n(U) = 30$
Set of students who like football = $F$, $\therefore n(F) = 20$
Set of students who like cricket = $C$, $\therefore n(C) = 15$
Set of students who like both games = $F \cap C$, $\therefore n(F \cap C) = 10$
Venn Diagram Explanation:
In the Venn diagram, the rectangle represents the universal set $U$. Inside $U$, two intersecting circles $F$ and $C$ divide the region into four disjoint regions ($r_1, r_2, r_3, r_4$):
- Students who like only football = $r_1$
- Students who like both games = $r_2$
- Students who like only cricket = $r_3$
- Students who like neither game = $r_4$
Calculations:
Students who like both games, $r_2 = 10$
Students who like only football, $r_1 = n(F) – r_2 = 20 – 10 = 10$
Students who like only cricket, $r_3 = n(C) – r_2 = 15 – 10 = 5$
Therefore, students who like at least one game, $n(F \cup C) = r_1 + r_2 + r_3 = 10 + 10 + 5 = 25$
Hence, the number of students who like neither game, $r_4 = n(U) – n(F \cup C) = 30 – 25 = 5$
Answer:$5$ students do not like either game.
12. Of $100$ students $65$ passed in Bangla, $48$ in both Bangla and English, and $15$ students failed in both subjects.
(1) If $(1, x + y) = (2x – y, 5)$, find $(x, y)$.
Solution: By the definition of ordered pairs,
$2x – y = 1 \text{ — (i)}$
and
$x + y = 5 \text{ — (ii)}$
Adding equations (i) and (ii), we get
$(2x – y) + (x + y) = 1 + 5$
$\implies 3x = 6$
$\implies x = 2$
Substituting $x = 2$ into equation (ii):
$2 + y = 5 \implies y = 3$
Answer:$(x, y) = (2, 3)$
(b) Determine the number of students who passed only in Bangla or passed only in English.
Solution: Let,
Set of total examinees = $U$, $\therefore n(U) = 100$
Set of students who passed in Bangla = $B$, $\therefore n(B) = 65$
Set of students who passed in both subjects = $B \cap E$, $\therefore n(B \cap E) = 48$
Number of students who failed in both subjects = $15$
- Passed only in Bangla:
$n(\text{only Bangla}) = n(B) – n(B \cap E) = 65 – 48 = 17$ students. - Passed only in English:
Number of students who passed in at least one or both subjects $= n(U) – \text{Failed in both} = 100 – 15 = 85$ students.
We know,
$n(B \cup E) = n(\text{only Bangla}) + n(\text{both subjects}) + n(\text{only English})$
$\implies 85 = 17 + 48 + n(\text{only English})$
$\implies 85 = 65 + n(\text{only English})$
$\implies n(\text{only English}) = 85 – 65 = 20$ students.
Answer:$17$ students passed only in Bangla and $20$ students passed only in English.
(c) Determine union of the sets of prime factors of the number of students passed and failed in both subjects.
Solution: Here,
Number of students who passed in both subjects = $48$
Number of students who failed in both subjects = $15$
Prime factors of $48$:
$48 = 2 \times 2 \times 2 \times 2 \times 3$
Let the set of prime factors of $48$ be $P = \{2, 3\}$
Prime factors of $15$:
$15 = 3 \times 5$
Let the set of prime factors of $15$ be $Q = \{3, 5\}$
The required union set ($P \cup Q$):
$P \cup Q = \{2, 3\} \cup \{3, 5\} = \{2, 3, 5\}$
Answer:$\{2, 3, 5\}$